Question 1172580
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Let *[tex \Large r_b] the speed of the bus and *[tex \Large r_t] the speed of the train.


Distance equals rate times time, which is to say time is equal to distance divided by rate.  Hence, *[tex \Large \frac{200}{r_b}] is the time it takes the bus to go 200 km, and *[tex \Large \frac{200}{r_t}] is the time it take the train to go 200 km.  We know these two times add up to 6.5 hours, hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{200}{r_b}\ +\ \frac{200}{r_t}\ =\ 6.5]


Similarly,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{300}{r_b}\ +\ \frac{100}{r_t}\ =\ 7.25]


Multiply the second equation by -2, and add the two equations to get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -\frac{400}{r_b}\ =\ -8]


Solve for *[tex \Large r_b].  The rest is just arithmetic.

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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